Minimize sum of absolute values of A[i] – (B + i) for a given array
Given an array arr[ ] of size N, the task is to find the minimum possible value of the expression abs(arr[1] – (b + 1)) + abs(arr[2] – (b + 2)) . . . abs(arr[N] – (b + N)), where b is an independent integer.
Input: arr[ ] = { 2, 2, 3, 5, 5 }
Output: 2
Explanation: Considering b = 0: The value of the expression is abs(2 – (0 + 1)) + abs(2 – (0 + 2)) + abs(3 – (0 + 3)) + abs(5 – (0 + 4)) + abs(5 – (0 + 5)) = 1 + 0 + 0 + 1 + 0 = 2
Therefore, the minimum possible value for the expression is 2.Input: arr[ ] = { 6, 5, 4, 3, 2, 1 }
Output: 18
Approach: Considering B[i] = A[i] − i, the problem is to reduces to minimize the sum of abs (B[i] − b). It can be observed that it is best to have b as the median of the modified array B[]. So, after sorting array B[], the problem can be solved following the steps given below.
- Traverse the array arr[ ] and decrease every element by their index (i + 1).
- Sort the array in ascending order.
- Now, choose b as the median of arr[ ], say b = arr[N/2].
- Initialize a variable, say ans as 0, to store the minimum possible value of the expression.
- Traverse the array again and update ans as abs(arr[i] – b).
- Return the value of ans.
Below is the implementation of the above approach.
C++
// C++ program for above approach #include <bits/stdc++.h> using namespace std; // Function to calculate minimum // possible sum of all (arr[i] - b + i) int MinSum( int arr[], int N) { // Modify the array for ( int i = 0; i < N; i++) { arr[i] = arr[i] - (i + 1); } // Sort the array sort(arr, arr + N); // Calculate median int b = arr[N / 2]; // Stores the required answer int ans = 0; for ( int i = 0; i < N; i++) { // Update answer ans += abs (arr[i] - b); } // Return the answer return ans; } // Driver Code int main() { // Given Input int arr[] = { 2, 2, 3, 5, 5 }; int N = sizeof (arr) / sizeof ( int ); // Function Call int ans = MinSum(arr, N); cout << ans << "\n" ; return 0; } |
Java
// Java program for above approach // Function to calculate minimum // possible sum of all (arr[i] - b + i) import java.util.*; class GFG{ static int MinSum( int arr[], int N) { // Modify the array for ( int i = 0 ; i < N; i++) { arr[i] = arr[i] - (i + 1 ); } // Sort the array Arrays.sort(arr); // Calculate median int b = arr[N / 2 ]; // Stores the required answer int ans = 0 ; for ( int i = 0 ; i < N; i++) { // Update answer ans += Math.abs(arr[i] - b); } // Return the answer return ans; } // Driver Code public static void main(String[] args) { // Given Input int arr[] = { 2 , 2 , 3 , 5 , 5 }; int N = arr.length; // Function Call int ans = MinSum(arr, N); System.out.print(ans); } } // This code is contributed by shivanisinghss2110 |
Python3
# Java program for above approach # Function to calculate minimum # possible sum of all (arr[i] - b + i) def MinSum(arr, N): # Modify the array for i in range (N): arr[i] - = (i + 1 ) # sort the array arr.sort() # calculate median b = arr[N / / 2 ] # Stores the required answer ans = 0 for i in range (N): # Update answer ans + = abs (arr[i] - b) # Return the answer return ans # Driver code arr = [ 2 , 2 , 3 , 5 , 5 ] N = len (arr) print (MinSum(arr, N)) # This code is contributed by Parth Manchanda |
C#
// C# program for above approach using System; class GFG{ // Function to calculate minimum // possible sum of all (arr[i] - b + i) static int MinSum( int []arr, int N) { // Modify the array for ( int i = 0; i < N; i++) { arr[i] = arr[i] - (i + 1); } // Sort the array Array.Sort(arr); // Calculate median int b = arr[N / 2]; // Stores the required answer int ans = 0; for ( int i = 0; i < N; i++) { // Update answer ans += Math.Abs(arr[i] - b); } // Return the answer return ans; } // Driver Code static void Main() { // Given Input int []arr = { 2, 2, 3, 5, 5 }; int N = arr.Length; // Function Call int ans = MinSum(arr, N); Console.Write(ans); } } // This code is contributed by SoumikMondal |
Javascript
<script> // JavaScript Program for the above approach // Function to calculate minimum // possible sum of all (arr[i] - b + i) function MinSum(arr, N) { // Modify the array for (let i = 0; i < N; i++) { arr[i] = arr[i] - (i + 1); } // Sort the array arr.sort( function (a, b) { return a - b }); // Calculate median let b = arr[Math.floor(N / 2)]; // Stores the required answer let ans = 0; for (let i = 0; i < N; i++) { // Update answer ans += Math.abs(arr[i] - b); } // Return the answer return ans; } // Driver Code // Given Input let arr = [2, 2, 3, 5, 5]; let N = arr.length; // Function Call let ans = MinSum(arr, N); document.write(ans + "<br>" ); // This code is contributed by Potta Lokesh </script> |
2
Time Complexity: O(N*logN)
Auxiliary Space: O(1)
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